The sequence 1,11,111,... and the prime factorization of its elements I have been recently investigating the sequence 1,11,111,... I found, contrary to my initial preconception, that the elements of the sequence can have a very interesting multiplicative structure. There are for example elements of the sequence that are divisible by primes like 7 or 2003.
What I am interested in is for what numbers, other than 2 and 5 can we say that they divide no element of the sequence?
 A: In fact, every number coprime with $10$ (that is, those that aren't integer multiples of $2$ and/or $5$) divides some element of that sequence. See this question.
On the other hand, it is immediately clear that no even number or integer multiple of $5$ can divide an element of that sequence.
A: As you can see in the answers to this question, a number has a multiple of the form $111...1$ if  the number is not divisible by $2$ and $5$ (i.e. relatively prime to 10). 
Conversely, if a number has a multiple of the form $111.11$ its multiple is not divisible by $2$ or $5$ since the last digit is $1$. Thus the number is  relatively prime to $2$ and $5$.
Conclusion: A number $n$ has a multiple of the form $11111....1$ if and only if the number is relatively prime to $10$.
P.S. Another interesting property. For any prime $p \neq 2,5$ it follows from Fermat Little Theorem that 
$$p|10^{p-1}-1 \,.$$
From here it follows immediately that for any prime $p \neq 2,3,5$, $p|111...1$, where there are exactly $p-1$ ones...
A: Indeed, there are such nice properties. To start, consider the following exercise from The Art and Craft of Problem Solving by Paul Zeitz. 
Example 1.2: There is an element in the sequence $ 7, 77, 777, \cdots $ that is divisible by $2003$. 
Proof: We prove that even a stronger statement is true, in fact, one of the first $2003$ elements of the sequence is divisible by $2003$. Let us assume that the contrary is true. Then take the first $2003$ elements of the sequence and divide each of them by $2003$. As none of them is divisible by $200$, they will have have a remainder that is at least $1$ and at most $2002$. As there are $2003$ remainder (one for each of the first $2003$ elements of the sequence), and only $2002$ possible values for these remainders, it follows by the Pigeonhole Principle that there are two elements out of the first $2003$ that have the same remainder. Let us say that the $i$th and $j$th elements of the sequence $a_i$ and $a_j$, have this property, and let $ i < j $. Consider the following difference: $$ \underbrace {777 \cdots 7}_{j \, \text{digits}} - \underbrace {77 \cdots 7}_{i \, \text{digits}} = \underbrace {7 \cdots 7}_{j-i \, \text{sevens}}\underbrace{0 \cdots 0}_{i \, \text{zeroes}}. $$As $a_i$ and $a_j$ have the same remainder when divided by $2003$, there exist non-negative integers $k_i$, $k_j$, and $r$ so that $ r \le 2002 $, and $ a_i = 2003k_i + r $ and $ a_j = 2003k_j + r $. This shows that $ a_j - a_i = 2003 \cdot (k_j - k_i) $, so in particular, $a_j-a_i$ is divisible by $2003$. 
This is nice, but we need to show that there is an element in our sequence that is divisible by $2003$, and $a_j-a_i$ is not an element in our sequence. However, the centered text above is very useful. 
Indeed, $a_j-a_i$ consists of $j-i$ digits equal to $7$, when $i$ digits equal to $0$. In other words, $$ a_j - a_i = a_{j-1} \cdot 10^i, $$ and the proof follows as $ 10^i $ is relatively primes to $2003$, so $a_{j-1}$ must be divisible by $2003$. 
$ \blacksquare $ 
Can you generalize? Also, see here. 
